128 Actions

How to “read” the temperature of an abstract system?forget what I wrote about f. It is confusing and wrong. In this formulation the free energy is given by $f=-k_bT\log Z=k_bT\log \sum_i e^{-\beta E_i}$, and the mean energy is calculated by averaging $\langle E\rangle=Z^{-1}\sum E_i e^{-\beta E_i}=-\partial_\beta \log(Z)$

How to “read” the temperature of an abstract system?The point is showing that coupling your system to an external bath results in a probability that goes like $e^{-\beta E}$, for some $\beta$. This is te definition of $T$, and is crux of the matter. Until you've shown this you don't know that "T will just take its value" from the bath, because $T$ is not even defined.

How to “read” the temperature of an abstract system?Systems flow towards probable states, that is - states that have a large number of micro-states. The (log of the) number of micro states is the entropy. So systems "want" to be in (="go to") states with high entropy, but that usually means states with high energy. The trade-off between entropy and energy is exactly the temperature. In evaporating liquid, for example, the vapor state has a higher entropy, but also a higher energy. Therefore, at low $T$ the system is liquid, but when $T$ is high enough the system prefers to be at a higher energetic state (="pay") because its entropy is higher.

How to “read” the temperature of an abstract system?You wrote the answer. This is the definition of $T$. In the Ising model. for example, there is no sense in talking about "average kinetic energy of the spins" or anything of that sort. I'd say that a good way to think about the temperature in this case is "how far above the ground state can I go", which is roughly equivalent to "how much energy can I pay in order to buy some entropy"?

Why doesn't phase space contain acceleration/forces?This is wrong - it's only good for an infinitesimal $\delta t$ later (i.e. to first order in $\delta t$). If you want to know the trajectory you need to know the accelerations. But these can be calculated from the position and velocity.

Jan11

comment

Do all closed systems, only considering kinematic/mechanical principles, exhibit time reversal symmetry?Strictly speaking, in classical mechanics the answer is that the dynamics are fully reversible. However, In the real world you are bounded, even theoretically, by the uncertainty principle, not to mention the outrageous impossibleness of reconstructing the the system with reversed velocities. Also, many-body dynamics are generically chaotic, and infinitesimal deviations of initial conditions will result in a significantly different evolution.